Infinitesimal transformations and Poisson brackets

  • #1
Hello, I want to understand how bracket operations in general are related to symmetry and infinitesimal transformations (in hindsight of quantumfieldtheory), so I calculated an example with a particle that is moving on a circle with a generic potential.
(I used simple polar coordinates in two dimensions)

[itex]H(r,p_{r})= \frac{p^{2}_{r}}{2m}+V(r)[/itex]
[itex]H(\phi, p_{\phi})=\frac{p^{2}_{\phi}}{2mr^{2}}+V(\phi)[/itex]

Now I know that if you take the Poisson bracket with the Hamiltonian you just get the infinitesimal transformation in time right? So
[itex]\{r,H(r,p_{r})\}= \frac{p_{r}}{m}[/itex]
[itex]\{\phi,H(\phi,p_{\phi})\}= \frac{p_{\phi}}{mr^{2}}[/itex]

But what if I want to do an infinitesimal transformation of the [itex]r[/itex] and [itex]\phi[/itex] coordinates? I know that the generator of translations is just the momentum, and that the generator of rotations is angular momentum. How would I do that with the Poisson bracket in this case? And for example when I do an infinitesimal transformation with the radius [itex]r[/itex], what does that mean? Is it that the radius is infinitesimally transformed, or is it more like a global translation where the whole system is somehow translated? Similarly with the angle [itex]\phi[/itex], is it that the angle is locally changed, or is it that the "whole" system is rotated?
 

Answers and Replies

  • #2
4
0
Now I know that if you take the Poisson bracket with the Hamiltonian you just get the infinitesimal transformation in time right? So
Are you referring to this formula
$$\frac{df}{dt} = \{f, H\} + \frac{\partial f}{\partial t} $$
As for infinitesimal transformation, it is just the radius ##r## changes. For example, we may have a infinitesimal canonical transformation
$$Q = q + \alpha G(q,P)$$
where ##\alpha## is infinitesimally small and ##G## is the generator.
 

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